Question

реши систему уравнений используя метод подстановки$\left \{ {{x+y=3} \atop {x^2+y^2=29}} \right.$

Answer 1

$\left \{ {{x+y=3} \atop {x^2+y^2=29}} \right. =>x=3-y\\\\(3-y)^2+y^2=29\\9-6y+y^2+y^2=29\\2y^2-6y-20=0 \ (:2)\\y^2-3y-10=0\\y^2+2y-5x-10=0\\y(y+2)-5(y+2)=0\\(y+2)(y-5)=0\\y_1=-2 \ \ \ \ \ \ \ \ y_2=5\\\\\left \{ {{x=3+2} \atop {x=3-5}} \right. =>\left \{ {{x_1=5} \atop {x_2=-2}} \right.$

ОТВЕТ:

$x_1=5 \ \ \ \ \ x_2=-2\\y_1=-2 \ \ \ \ \ y_2=5$